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A block of mass 1kg is hanging vetically from a string of length 1m and Mass/length =0.001 kg/m. A small pulse is generated at its lower end. The Pulse reaches the top end in A. 0.2 sec B. 0.1 sec C. 0.02 sec D. 0.01 sec?
Verified Answer
A block of mass 1kg is hanging vetically from a string of length 1m an...
Correct option (D) 0.01 sec
Explanation:
Explanation:
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Most Upvoted Answer
A block of mass 1kg is hanging vetically from a string of length 1m an...
0.01 sec is correct, dude...
The question is there in Fundamental Physics Standard XI.
The question is there in Fundamental Physics Standard XI.
Community Answer
A block of mass 1kg is hanging vetically from a string of length 1m an...
Solution:
Given:
Mass of the block, m = 1 kg
Length of the string, L = 1 m
Mass per unit length, μ = 0.001 kg/m
To find:
The time taken for the pulse to reach the top end of the string.
Analysis:
When a small pulse is generated at the lower end of the string, it will propagate through the string as a wave. The wave speed, v, can be calculated using the formula:
v = √(T/μ)
where T is the tension in the string.
The tension in the string can be calculated by considering the forces acting on the block. At equilibrium, the weight of the block is balanced by the tension in the string. Therefore:
T = mg
where g is the acceleration due to gravity.
Once the wave speed is known, the time taken for the pulse to reach the top end of the string can be calculated using the formula:
Time = Distance / Velocity
The distance traveled by the pulse is equal to the length of the string, L.
Calculation:
1. Calculating the tension in the string:
T = mg
= 1 kg * 9.8 m/s^2
= 9.8 N
2. Calculating the wave speed:
v = √(T/μ)
= √(9.8 N / 0.001 kg/m)
= √9800 m/s
≈ 99 m/s
3. Calculating the time taken for the pulse to reach the top end:
Time = Distance / Velocity
= 1 m / 99 m/s
≈ 0.01 s
Therefore, the time taken for the pulse to reach the top end of the string is approximately 0.01 seconds.
Answer: D. 0.01 sec
Given:
Mass of the block, m = 1 kg
Length of the string, L = 1 m
Mass per unit length, μ = 0.001 kg/m
To find:
The time taken for the pulse to reach the top end of the string.
Analysis:
When a small pulse is generated at the lower end of the string, it will propagate through the string as a wave. The wave speed, v, can be calculated using the formula:
v = √(T/μ)
where T is the tension in the string.
The tension in the string can be calculated by considering the forces acting on the block. At equilibrium, the weight of the block is balanced by the tension in the string. Therefore:
T = mg
where g is the acceleration due to gravity.
Once the wave speed is known, the time taken for the pulse to reach the top end of the string can be calculated using the formula:
Time = Distance / Velocity
The distance traveled by the pulse is equal to the length of the string, L.
Calculation:
1. Calculating the tension in the string:
T = mg
= 1 kg * 9.8 m/s^2
= 9.8 N
2. Calculating the wave speed:
v = √(T/μ)
= √(9.8 N / 0.001 kg/m)
= √9800 m/s
≈ 99 m/s
3. Calculating the time taken for the pulse to reach the top end:
Time = Distance / Velocity
= 1 m / 99 m/s
≈ 0.01 s
Therefore, the time taken for the pulse to reach the top end of the string is approximately 0.01 seconds.
Answer: D. 0.01 sec
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A block of mass 1kg is hanging vetically from a string of length 1m and Mass/length =0.001 kg/m. A small pulse is generated at its lower end. The Pulse reaches the top end in A. 0.2 sec B. 0.1 sec C. 0.02 sec D. 0.01 sec?
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A block of mass 1kg is hanging vetically from a string of length 1m and Mass/length =0.001 kg/m. A small pulse is generated at its lower end. The Pulse reaches the top end in A. 0.2 sec B. 0.1 sec C. 0.02 sec D. 0.01 sec? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A block of mass 1kg is hanging vetically from a string of length 1m and Mass/length =0.001 kg/m. A small pulse is generated at its lower end. The Pulse reaches the top end in A. 0.2 sec B. 0.1 sec C. 0.02 sec D. 0.01 sec? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A block of mass 1kg is hanging vetically from a string of length 1m and Mass/length =0.001 kg/m. A small pulse is generated at its lower end. The Pulse reaches the top end in A. 0.2 sec B. 0.1 sec C. 0.02 sec D. 0.01 sec?.
A block of mass 1kg is hanging vetically from a string of length 1m and Mass/length =0.001 kg/m. A small pulse is generated at its lower end. The Pulse reaches the top end in A. 0.2 sec B. 0.1 sec C. 0.02 sec D. 0.01 sec? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A block of mass 1kg is hanging vetically from a string of length 1m and Mass/length =0.001 kg/m. A small pulse is generated at its lower end. The Pulse reaches the top end in A. 0.2 sec B. 0.1 sec C. 0.02 sec D. 0.01 sec? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A block of mass 1kg is hanging vetically from a string of length 1m and Mass/length =0.001 kg/m. A small pulse is generated at its lower end. The Pulse reaches the top end in A. 0.2 sec B. 0.1 sec C. 0.02 sec D. 0.01 sec?.
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